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Tatiana [17]
2 years ago
12

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N

and points away from q1 and q3. What is the value (sign and magnitude) of the charge q3?
Physics
1 answer:
Dmitriy789 [7]2 years ago
4 0

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

r=\sqrt{(r_{1})^2+(r_{2})^2}

Put the value into the formula

r=\sqrt{(9.8)^2+(2.1)^2}

r=10.02\ cm

We need to calculate the force

Using formula of force

F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}

Force F₁₂,

F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}

F_{12}=13.33\ N

F_{21}=-13.33\ N

Force F₂₃,

F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}

We need to calculate the value of third charge

F_{net}=F_{21}+F_{23}

4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}

q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}

q_{3}=7.99\times10^{-7}\ C

q_{3}=0.8\times10^{-6}\ C

Hence, The value of third charge is 0.8μC.

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Answer:

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Explanation:

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The area of ​​the sphere is

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Since the amount of radiation emitted by the star is constant, we can write this expression for the position of the two planets

               P = I₁ A₁ = I₂ A₂

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 Suppose index 1 corresponds to the nearest planet,

            r2 = 2 r₁

            I₁ / I₂ = r₁² / r₂²

            I₁ / I₂ = r₁² / (2r₁)²

            I₁ / I₂ = ¼

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The nearest plant (A) receives 4 times more radiation from the farthest plant

7 0
3 years ago
A cube made Of an unknown material has a height of 9 symeters the mass of this cube is 3645 g. calculate the density of this cub
kherson [118]

Answer: 5 \frac{g}{cm^{3}}

Explanation:

The density \rho of a material is given by:

\rho=\frac{m}{V} (1)

Where:

m=3645 g is the mass of the cube

V is the volume of the cube

Now, the volume of a cube is equal to the length L of its edge to the power of 3:

V=L^{3} (2)

If we know L=9 cm, the volume of this cube is:

V=(9 cm)^{3}=729 cm^{3} (3)

Substituting (3) in (1):

\rho=\frac{3645 g}{729 cm^{3}} (4)

\rho=5 \frac{g}{cm^{3}} This is the density of the cube

8 0
3 years ago
Three boys of equal strength try to break a rope (and fail) by tying one end to a fence post and tugging on the other end. Three
podryga [215]

The best choice would be C. Keep one end of the rope tied to the post and have all six boys tug on the other end.

Reason:

A is not correct since in the first case which is B, untying the rope from the post and have 3 boys on each end of it tug, will provide equal or a near equal amount of tension since the boys are all within the same strength, however there can be variables that will affect the tension.

That means B would be true according to A, and then for C, having all six boys pull on the other end of the rope would provide greater tension, since if 3 boys of the same strength range can't break the rope when its tied to the fence, we add the other 3 boys of the same strength.

However, if 3 boys are on each end of the rope and of the same strength then the rope will not break since there is an equal amount of net force exerted on the rope.

We know that the fence can withstand the strength of 3 boys but if we add the other 3 boys then it could provide us with a different outcome.

3B < F

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So, therefore, our best choice would be C since A and B is incorrect.

4 0
3 years ago
A boat is able to move through still water at 20 m/s. It makes a round trip to a town 10 km upstream. If the river flows at 5m/s
Nataly_w [17]

Answer:

The time required for this round trip is 1066.66 s

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time = d / v

The boat made two journeys which formed the round trip.

During the first part of the journey, the boat moves upstream in opposite direction to the flow of the river and the resultant velocity is calculated as;

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Time for this journey, = 10000/ 15 = 666.66 s

During the second part of the journey, the boat moves downstream in the same direction to the flow of the river and the resultant velocity is calculated as;

Resultant velocity = 20 + 5 = 25 m/s

Time for this journey, = 10000/ 25 = 400 s

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vagabundo [1.1K]

The original frequency of horn of Car A is 1071 Hz.

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Here vo is the observer velocity and vs is the velocity of the source. So Vo = 15 m/s as car B is the observer and Vs = 35 m/s as car A is the source. And f is the frequency of sound wave at source that is car A.

Similarly, the doppler shift in frequency is the frequency of sound heard by car B which is f' = 1140 Hz. And v is the speed of sound that is v = 343 m/s

1140 = \frac{343-15}{343-35}*f= 1.0649 *f

f = 1140/1.0649= 1071 Hz.

Thus, the original frequency of horn of Car A is 1071 Hz.

3 0
3 years ago
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